Question

find the de Broglie's wavelength of the electron in the first Bohr's orbit ?

Answer 1

De Broglie wavelength
$\gamma = \frac{h}{mv} \\ \\ where \\ \gamma = wavelength \: of \: radiation \\ h = planks \: constant \\ m = mass \: of \: electron \\ v = velocity \: of \: electron$

A/c to Bohr's

velocity of electron in first orbit

$v = v_{o} \times \frac{z}{n} \\ \\ where \\ v = velocity \: of \: electron \: \\ v_{o} = 2.18 \times {10}^{6} m {s}^{ - 1} \\ z = atomic \: no. \\ n = no. \: of \: orbit$
Now let here is the atom of Hydrogen
then,
d
$v = v_{o} \times \frac{z}{n} \\ \\ v = 2.18 \times {10}^{6} \times \frac{1}{1} \\ \\ :. \: v = 2.18 \times {10}^{6}$

Now, In De Broglie wavelength

$\gamma = \frac{h}{mv} \\ \\ \gamma = \frac{6.6 \times {10}^{ - 34} }{9.1 \times {10}^{ - 31} \times 2.18 \times {10}^{6} } \\ \\ \gamma = \frac{6.6 \times {10}^{ (- 34 + 31 - 6)} }{9.1 \times 2.18} \\ \\ \gamma = 0.3326948281 \times {10}^{ - 9} m$
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Answer 2

Explanation:

here is your answer the motion not pasted